在将 XML 映射到 Java Bean 方面需要帮助

Question

我有一个如下的 XML ...

<employeeId>323</employeeId>
<name>Samuel DCosta</name>
<department>
    <departmentId>2</departmentId>
    <name>Accounts</name>
</department>
<salary>11290</salary>

我想将这些值映射到我拥有的 Java Bean .... XML 中的键与 bean 中成员的名称匹配 ... 有人告诉我是否有简单的方法请在 Java 中执行此操作 .... 欢迎使用工具或组件 ...

部门....

import java.io.Serializable;

public class Department implements Serializable
{
private Long departmentId;

private String name;

@Override
public String toString()
{
    return "Department [departmentId=" + departmentId + ", name=" + name + "]";
}

public Long getDepartmentId()
{
    return departmentId;
}

public void setDepartmentId(Long departmentId)
{
    this.departmentId = departmentId;
}

public String getName()
{
    return name;
}

public void setName(String name)
{
    this.name = name;
}
}

员工.....

import java.io.Serializable;

public class Employee implements Serializable
{
private Long employeeId;

private String name;

private Department department;

private Integer salary;

@Override
public String toString()
{
    return "Employee [employeeId=" + employeeId + ", name=" + name + ", department=" + department + ", salary="
            + salary + "]";
}

public Long getEmployeeId()
{
    return employeeId;
}

public void setEmployeeId(Long employeeId)
{
    this.employeeId = employeeId;
}

public String getName()
{
    return name;
}

public void setName(String name)
{
    this.name = name;
}

public Department getDepartment()
{
    return department;
}

public void setDepartment(Department department)
{
    this.department = department;
}

public Integer getSalary()
{
    return salary;
}

public void setSalary(Integer salary)
{
    this.salary = salary;
}
}

Answer 1

您可以使用JAX-B

用于 XML 绑定的 Java 架构 (JAXB) 提供了一种快速且 绑定 XML 模式和 Java 表示的便捷方式，使其 Java 开发人员易于合并 XML 数据和处理 Java 应用程序中的函数。作为此过程的一部分，JAXB 提供 将 XML 实例文档解组（读取）成 Java 的方法 内容树，然后编组（写入）Java 内容树 到 XML 实例文档中。 JAXB 还提供了一种生成 XML 的方法 Java 对象的架构

Answer 2

首先，将 JAXB 注解添加到您的 bean 类中：

    @XmlRootElement
    public class Employee {
    private Long employeeId;
    private String name;
    private Department department;

    private Integer salary;
    //getter and setter omitted here
   }

@XmlRootElement
public class Department {
    private Long departmentId;
    private String name;
     //getter and setter omitted here
}

这是我用于测试的“employee.xml”文件：

<employee>
    <employeeId>323</employeeId>
    <name>Samuel DCosta</name>
    <department>
        <departmentId>2</departmentId>
        <name>Accounts</name>
    </department>
    <salary>11290</salary>
</employee>

然后就可以像这样读取XML文件了

public class EmployeeReader {
public static <T> T fromXML(Reader reader,Class<T> type) throws JAXBException {
    JAXBContext context=JAXBContext.newInstance(type);
    Unmarshaller unmarshaller=context.createUnmarshaller();
    return (T)unmarshaller.unmarshal(reader);
}

    public static void main(String[] args)
    {
        Reader reader=null;
        try
        {
            reader=new FileReader("employee.xml");
            Employee employee= EmployeeReader.fromXML(reader,Employee.class);
            System.out.println(employee.getName());
            System.out.println(employee.getDepartment().getName());
        }catch (Exception e)
        {
            e.printStackTrace();
        }finally {
        IOUtils.closeQuietly(reader);
    }
    }
}

Answer 3

我在测试代码中使用了您的原始 XML 和 2 个 Java Bean——Employee 和 Department

import java.util.ArrayList;
import java.util.List;

import cjm.component.cb.object.ToObject;

public class EmployeeConversion
{
public static void main(String[] args)
{
    try
    {
        String xml = "<employeeId>323</employeeId><name>Samuel DCosta</name><department><departmentId>2</departmentId><name>Accounts</name></department><salary>11290</salary>";

        List<Object> objectList = new ArrayList<Object>();

        objectList.add(new Employee());
        objectList.add(new Department());         // adding all the nested objects within the Employee bean into this list

        Employee employee = (Employee) new ToObject().convertToObject(xml, new Employee(), objectList); // this will carry out the mapping to the bean

        System.out.println(employee);
    }
    catch (Exception e)
    {
        e.printStackTrace();
    }
}
}

输出

-------- XML Detected -------- 
-------- XML Detected -------- 
-------- Map created Successfully -------- 
-------- Object created Successfully -------- 
Employee [employeeId=323, name=Samuel DCosta, department=Department [departmentId=2, name=Accounts], salary=11290]

您将需要转换框（适用于 v1.1） http://capsulesforthejavamind.blogspot.in/2015/01/conversion-box.html

告诉我结果！！！ :)

Answer 4

您可以为您提供的 xml 生成一个 xsd 文件，然后使用 JAXB 使用 eclipse 自动生成与架构对应的 java 类。 对于自动生成 java 类，您可以参考以下链接： How can I get the "Eclipse >Generate>Jaxb classes" option back?